Computing the Supremum of a Family

Question

I am trying to compute (or bound) the supremum $\sup_{n\in \mathbf{Z}} n^p x^n$ for fixed $p<2\in \mathbf{R}$. It should be noted that $x\in [0,1]$. My professor suggested that I bound this by $$ \sup_{n\in \mathbf{N}}n^px^n\le \sup_{t\in \mathbf{R}} t^px^t.$$ From here, I am trying to compute a bound for the supremum over real numbers $t$. It looks like this can be achieved using some calculus trickery. Namely, set $f(t)=t^pe^{(\ln x) \cdot t}$. Then, we have that $$ f'(t)=pt^{p-1}e^{(\ln x)\cdot t}+t^p \ln xe^{(\ln x)\cdot t}=x^tt^{p-1}(p+t\ln x).$$ Setting this derivative to $0$, we have that $$ x^t t^{p-1}(p+t\ln x)=0.$$ Now, either $t=0$, $x=0$, or $(p+t\ln x)=0$. Clearly the $t=0$ case is not a maximum, and the $x=0$ case is trivial because for $x$, the family has value $0$ for all $t$. So, the only other zero I can find is $$ t=\frac{-p}{\ln x}.$$ I'm not exactly sure what to do with this information. If anyone could give me a hint or suggestion, I would greatly appreciate it.

Answer 1

I'm assuming $n$ is natural, according to one of your comments. If $p \le 0$, you can bound $n^p x^n \le x^n \le 1$ and don't even need to pass to the real variable $t$. Therefore let's say $p > 0$.

Assume $p$ and $0 < x < 1$ are fixed; the boundary cases for $x$ are trivial. It suffices for your original purpose to take $t \ge 0$.

Notice that the sign of $f'(t)$ is dictated by the term $p + t \ln x$, which is linear in $t$. It therefore switches sign exactly once, from positive to negative (since $\ln x < 0$), at $t = t_0 = -p/\ln x > 0$, as you stated. Hence your supremum is $$f(t_0) = \left ( \frac{-p}{e \ln x} \right )^p.$$